1 / 67

Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Chapter 15 Solutions—9 out of 75 multiple choice Free Response—almost every year. Common Ion Effect. Common ion—an ion that is produced by more than one species in a solution Shifts equilibrium because concentration of common ion is increased. Example.

wilmet
Download Presentation

Applications of Aqueous Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Applications of Aqueous Equilibria Chapter 15 Solutions—9 out of 75 multiple choice Free Response—almost every year

  2. Common Ion Effect • Common ion—an ion that is produced by more than one species in a solution • Shifts equilibrium because concentration of common ion is increased

  3. Example AgCl(s)  Ag+(aq) + Cl(aq)

  4. Example • Nitrous acid (HNO2) and sodium nitrite (NaNO2) • Ka = [H+][NO2] / [HNO2] • Acid will dissociate less because the solution has nitrite ion present already

  5. Sample Problem • Calculate the [H+] and percent dissociation of a solution that is 1.00 M in HF and 1.00 M in NaF. The Ka of HF is 7.2 x 10-4.

  6. Buffers • Solutions that resist change in pH when either H+ or OH- are added • Contains a weak acid & its salt or a weak base & its salt • Takes advantage of common ion effect

  7. Key Points on Buffered Solutions 1.They are weak acids or bases containing a common ion. 2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

  8. Sample Problem • A buffer solution contains .50 M acetic acid and .50 M sodium acetate. What is the pH of the solution? (The Ka for acetic acid is 1.8 x 10-5.)

  9. Sample Problem • Calculate the pH of a solution that is .75 M in lactic acid and .25 M in sodium lactate. (Ka for lactic acid = 1.4 x 10-4.)

  10. Henderson-Hasselbalch Equation • Useful for calculating pH when the [A]/[HA] ratios are known.

  11. Sample Problem • What is the pH of a buffer solution that contains .50 M acetic acid and .50 M sodium acetate? Ka for acetic acid is 1.8 x 10-5.

  12. Sample Problem • Calculate the pH of a solution that is .75 M in lactic acid and .25 M in sodium lactate. (Ka for lactic acid = 1.4 x 10-4.)

  13. Sample problem • What is the pH of a buffer solution that contains .25 M NH3 (Kb = 1.8 x 10-5) and .40 M NH4Cl?

  14. pH Change in Buffered Solutions • Buffers contain relatively large amounts of weak acid and corresponding base. • Added H+ reacts to completion with the weak base. • Added OH- reacts to completion with weak acid

  15. pH Change in Buffered Solutions • The pH is determined by the ratio of the concentrations of the weak acid and weak base. • Think H-H equation.

  16. Sample Problem • How will the pH change when .010 mol of NaOH is added to 1.0 L of the acetic acid/sodium acetate buffer system? (Original pH was 4.74.)

  17. Sample Problem Calculate the pH change when .010 mol of gaseous HCl is added to 1.0 L of a buffer solution containing 5.00 M HC2H3O2 and 5.0 M NaC2H3O2. Ka = 1.8 x 10-5

  18. Buffering Capacity the amount of H+ or OH the buffer can absorb without a significant change in pH

  19. Titration (pH) Curve • A plot of pH of the solution being analyzed versus the amount of titrant added. • Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

  20. Strong Acid/Strong Base • Easiest type—acid & base dissociate completely and react completely • Initial—pH depends only on concentration • Add acid/base—do stoichiometry

  21. Strong Acid/Strong Base • pH depends on amount of substance remaining after reaction

  22. Sample Problem • You are titrating 50.00 mL of a .50 M NaOH solution with 1.0 M HCl. • What is the pH of the base solution before any acid has been added?

  23. Sample Problem • What is the pH of the base solution after 10.00 mL of acid have been added?

  24. Sample Problem • What is the pH of the solution after 12.5 mL of acid have been added?

  25. Sample Problem • What is the pH of the solution after 20.00 mL of acid have been added?

  26. Sample Problem • What is the pH of the solution after 25.00 mL of acid have been added?

  27. Sample Problem • What is the pH of the solution after 30.00 mL of acid have been added?

  28. Weak Acid/Strong Base • Initailly—find pH by doing a weak acid problem • After adding base—buffer problem—Use H-H equation • At midpoint—pH = pKa

  29. Weak Acid/Strong Base • At equivalence point—find pH by doing a weak base problem for conj. base • After end point—pH depends on amount of excess base

  30. Sample Problem • Titration of 50.0 mL of .10 M acetic acid (Ka = 1.8 x 10-5) with .10 M NaOH. • What is the initial pH of the acetic acid solution?

  31. Sample Problem • What is the pH after 15.0 mL of base have been added?

  32. Sample Problem • What is the pH after 25.0 mL of base have been added?

  33. Sample Problem • What is the pH after 35.0 mL of base have been added?

  34. Sample Problem • What is the pH at the endpoint?

  35. Sample Problem • What is the pH after 60.0 mL of base have been added?

  36. Weak Base/Strong Acid • Initailly—find pH by doing a weak base problem • After adding acid—buffer problem—Use H-H equation • At midpoint—pH = pKa of conjugate acid

  37. Weak Base/Strong Acid • At equivalence point—find pH by doing a weak acid problem for conj. acid • After end point—pH depends on amount of excess acid

More Related